The group $H^1(K(\mathbb Z/4,1);\mathbb F_2)$ is one-dimensional, so it suffices to consider its nonzero class $x$. This class is represented by the quotient homomorphism $\mathbb Z/4\to\mathbb Z/2$, which lifts to the identity homomorphism with coefficients in $\mathbb Z/4$. Its <Bockstein homomorphism> for
$$
0\longrightarrow\mathbb Z/2\longrightarrow\mathbb Z/4
\longrightarrow\mathbb Z/2\longrightarrow0
$$
therefore vanishes. Since the first <Steenrod square> is this Bockstein and $\operatorname{Sq}^1x=x^2$ for every degree-one class,
$$
x^2=0.
$$
The zero degree-one class plainly has square zero as well.

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Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 165 / 4 / iii / Solution

A one-term summand $\mathbb Z[i]$ becomes a one-term $\mathbb F_p$ complex after reduction modulo $p$. Its <Bockstein homomorphism> vanishes, so it contributes one $\mathbb Z/p$ summand to <Bockstein homology> in degree $i$.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 114 / 1 / i / Solution

Because every $C_i$ is free abelian, reduction modulo $n$ gives a short exact sequence of <chain complex>[chain complexes]
$$
0\longrightarrow C_\bullet\xrightarrow{\ n\ }C_\bullet\xrightarrow{\rho}C_\bullet\otimes\mathbb Z/n\longrightarrow0.
$$
Its <long exact sequence in homology> is
$$
\cdots\to H_i(C)\xrightarrow{n}H_i(C)\xrightarrow{\rho_*}H_i(C\otimes\mathbb Z/n)\xrightarrow{\widehat\beta}H_{i-1}(C)\to\cdots.
$$
If $\bar z$ is a cycle modulo $n$, choose a lift $z\in C_i$. Then $dz=nw$ for some $w\in C_{i-1}$, and the <connecting homomorphism> is $\widehat\beta[\bar z]=[w]$.

The coefficient sequence
$$
0\longrightarrow\mathbb Z/n\xrightarrow{[a]\mapsto[na]}\mathbb Z/n^2\xrightarrow{[b]\mapsto[b]\bmod n}\mathbb Z/n\longrightarrow0
$$
similarly gives
$$
\cdots\to H_i(C\otimes\mathbb Z/n)\to H_i(C\otimes\mathbb Z/n^2)\to H_i(C\otimes\mathbb Z/n)\xrightarrow{\beta}H_{i-1}(C\otimes\mathbb Z/n)\to\cdots.
$$
At chain level, if $z$ lifts a mod-$n$ cycle and $dz=nw$ modulo $n^2$, then the <Bockstein homomorphism> is $\beta[\bar z]=[\bar w]$.

There is a morphism from the first short exact sequence to the second whose three vertical maps are reduction modulo $n$, reduction modulo $n^2$, and the identity on $C\otimes\mathbb Z/n$. Naturality of connecting homomorphisms gives
$$
\beta=\rho_*\widehat\beta.
$$
Exactness of the first long exact sequence says $\widehat\beta\rho_*=0$, and hence
$$
\beta^2=\rho_*\widehat\beta\rho_*\widehat\beta=0.
$$

For the standard cellular chain complex of <Real projective space> $\mathbb{RP}^3$, there is one copy of $\mathbb Z$ in degrees $0,1,2,3$, with $d_2=2$ and $d_1=d_3=0$. Modulo two all cellular differentials vanish, while the Bockstein is the identity from degree two to degree one and zero elsewhere. Therefore
$$
\beta H_i(C_*(\mathbb{RP}^3);2)\cong
\begin{cases}
\mathbb Z/2,&i=0,3,\\
0,&i=1,2.
\end{cases}
$$

Finally, Smith normal form decomposes a bounded chain complex of finitely generated free abelian groups, up to chain isomorphism and contractible summands, into one-term complexes $\mathbb Z[i]$ and two-term complexes
$$
0\longrightarrow\mathbb Z\xrightarrow{m}\mathbb Z\longrightarrow0
$$
in degrees $i+1$ and $i$. The former represents a free homology summand, and the latter represents $\mathbb Z/m$ in degree $i$. The two requested conclusions now follow from the next two parts.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 114 / 1 / Solution

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{title2=$\beta$}
{wiki}

The Bockstein homomorphism is the connecting homomorphism associated with a short exact sequence of coefficient groups. For $0\to\mathbb Z/n\to\mathbb Z/n^2\to\mathbb Z/n\to0$, it has degree $-1$ on homology and degree $+1$ on cohomology.


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 Incoming links: Bockstein homomorphism